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In the figure above, two security lights, L1 and L2 , are located 100

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In the figure above, two security lights, L1 and L2 , are located 100  [#permalink]

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New post 18 Feb 2018, 22:02
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In the figure above, two security lights, L1 and L2 , are located 100 feet apart. Each illuminates an area of radius 100 feet, and both are located 60 feet from a chain-link fence. What is the total length s of fence, in feet, illuminated by the two lights?

(A) 260

(B) 240

(C) 220

(D) 200

(E) 180

Attachment:
2018-02-19_1000.png
2018-02-19_1000.png [ 15.9 KiB | Viewed 911 times ]

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In the figure above, two security lights, L1 and L2 , are located 100  [#permalink]

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New post 18 Feb 2018, 23:35
Bunuel wrote:
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In the figure above, two security lights, L1 and L2 , are located 100 feet apart. Each illuminates an area of radius 100 feet, and both are located 60 feet from a chain-link fence. What is the total length s of fence, in feet, illuminated by the two lights?

(A) 260

(B) 240

(C) 220

(D) 200

(E) 180

Attachment:
2018-02-19_1000.png


100 . 80 . 60 triangle is formed wth the radius, fence and distance between fence n light source

three such triangles i.e length = 80+80+80 = 240

but some length isnt convered by three triangles thus >240

(a) imo
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Re: In the figure above, two security lights, L1 and L2 , are located 100  [#permalink]

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New post 19 Feb 2018, 00:47
Answer: A (260).

The lights make a 90 degree angle with the fence and hence the vertical distance is 60 feet. The lights cover 100 feet of radius , which means the triangle made is a pythagorus one with 100, 60 and uknown variable (calculate it to get 80). Thus, the light L1 covers a distance of 2(80) on fence and the L2 covers a little more than 80, making a total of more than 240.
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Re: In the figure above, two security lights, L1 and L2 , are located 100  [#permalink]

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New post 19 Feb 2018, 09:22
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The centers of both the circles are at a distance of 60 feet from the fence.

Since the maximum distance that the light will travel is 100 feet,
using the Pythagoras theorem the vertical distance can be calculated as \(\sqrt{100^2 - 60^2} = \sqrt{6400} = 80\)

But the small distance(marked in red) in the figure is not yet accounted for.
The total length(s) of the fence which is illuminated by the lights is a little over 240.

Therefore, Option A(260) is the only option available(> 240) and is our answer.
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Re: In the figure above, two security lights, L1 and L2 , are located 100  [#permalink]

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New post 20 Apr 2018, 04:35
The distance from L1 to L2 is 100.
If we drop a perpendicular from L1 to the fence at P and from L2 to the fence at Q, then the distance between P & Q will be 100. (L1 L2 Q P will be a rectangle)
100 . 80 . 60 triangle is formed with the radius, fence and distance between fence & light source.
So total distance = 80+100+80 = 260
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Re: In the figure above, two security lights, L1 and L2 , are located 100 &nbs [#permalink] 20 Apr 2018, 04:35
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In the figure above, two security lights, L1 and L2 , are located 100

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